Question: Solve for $x$. Enter the solutions from least to greatest. $(x - 10)^2 - 1 = 0$ $\text{lesser }x = $
Solution: $\begin{aligned} (x - 10)^2 - 1 &= 0 \\\\ (x-10)^2&=1 \\\\ \sqrt{(x-10)^2}&=\sqrt{1} \end{aligned}$ $\begin{aligned} x-10&=\pm1 \\\\ x&=\pm1+10 \\ \phantom{(x - 10)^2 - 1 }& \\ x=9&\text{ or }x=11 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= 9 \\\\ \text{greater } x &= 11 \end{aligned}$